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A061548 Numerator of probability that there is no error when average of n numbers is computed, assuming errors of +1, -1 are possible and they each occur with p = 1/4. 6

%I #35 Oct 27 2023 22:00:46

%S 1,3,35,231,6435,46189,676039,5014575,300540195,2268783825,

%T 34461632205,263012370465,8061900920775,61989816618513,

%U 956086325095055,7391536347803839,916312070471295267,7113260368810144185,110628135069209194801,861577581086657669325,26876802183334044115405

%N Numerator of probability that there is no error when average of n numbers is computed, assuming errors of +1, -1 are possible and they each occur with p = 1/4.

%H Indranil Ghosh, <a href="/A061548/b061548.txt">Table of n, a(n) for n = 0..500</a>

%H Robert M. Kozelka, <a href="http://www.jstor.org/stable/2322033">Grade Point Averages and the Central Limit Theorem</a>, American Mathematical Monthly. Nov. 1979 (86:9) pp. 773-7.

%F a(n) = numerator(binomial(2*n-1/2, -1/2)).

%F a(n) = numerator((4*n)!/(2^(4*n)*(2*n)!^2)). - _Johannes W. Meijer_, Jul 06 2009

%F a(n) = A001448(n)/A001316(n). - _Peter Luschny_, Mar 23 2014

%F a(n) is the numerator of the coefficient of power series in x around x=0 of sqrt(1 + sqrt(1 - x))/(sqrt(2)*sqrt(1 - x)). - _Karol A. Penson_, Apr 16 2018

%e For n=1, the binomial(2*n-1/2, -1/2) yields the term 3/8. The numerator of this term is 3, which is the second term of the sequence.

%p seq(numer(binomial(2*n-1/2, -1/2)), n=0..20);

%t Table[Numerator[(4*n) !/(2^(4*n)*(2*n) !^2) ], {n, 0, 20}] (* _Indranil Ghosh_, Mar 11 2017 *)

%t Table[Numerator[SeriesCoefficient[Series[(Sqrt[1 + Sqrt[1 - x]]/Sqrt[2 - 2* x]), {x, 0, n}], n]], {n, 0, 20}]. (* _Karol A. Penson_, Apr 16 2018 *)

%o (Sage)

%o def A061548(n): return binomial(4*n,2*n)/2^sum(n.digits(2))

%o [A061548(n) for n in (0..20)] # _Peter Luschny_, Mar 23 2014

%o (PARI) for(n=0, 20, print1(numerator((4*n)!/(2^(4*n)*(2*n)!^2)),", ")) \\ _Indranil Ghosh_, Mar 11 2017

%o (Python)

%o import math

%o from fractions import gcd

%o f = math.factorial

%o def A061548(n): return f(4*n) / gcd(f(4*n), (2**(4*n)*f(2*n)**2)) # _Indranil Ghosh_, Mar 11 2017

%Y Cf. A061549. Bisection of A001790.

%Y Equals 2*A001448(n)/ A117973(n). - _Johannes W. Meijer_, Jul 06 2009

%K nonn,frac,easy

%O 0,2

%A Leah Schmelzer (leah2002(AT)mit.edu), May 16 2001

%E More terms from _Asher Auel_, May 20 2001

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Last modified April 25 09:49 EDT 2024. Contains 371967 sequences. (Running on oeis4.)